In this section, we give some a priori estimates for a weak solution to (1.1). All the estimates hold for the weak solutions to (1.1) if we assume that the solutions exist. In the remainder of this paper, we denote \(B_{\rho }(x)\cap \Omega\) for \(x\in \Omega\), \(\rho >0\) by simply \(B_{\rho }(x)\) unless otherwise specified.
Proposition 3.1
(A priori estimate) Let r satisfy (1.3) and \(r>1\). Let \(u_0\in L_{\rho }^{r,\nu }(\Omega )\) and u be a \({L}^{r,\nu }(\Omega )\)- solution of (1.1) in \((0, T)\times \Omega\), where \(T>0\). There exists a positive constant \(\gamma _1\) such that, if
$$\begin{aligned} \rho ^{\frac{1}{p-1}-\frac{1}{n}} \sup _{0\le s\le T}\Vert u(s)\Vert _{L_{\rho }^{r,\nu }} \le \gamma _1 \end{aligned}$$
(3.1)
for some \(\rho >0\), then there exists a constant \(\mu >0\) depending only on p, r, n and \(\gamma _1\) such that
$$\begin{aligned} \sup _{0<s<t}\Vert u(s)\Vert _{L_{\rho }^{r,\nu }} \le C \Vert u_0\Vert _{ L_{\rho }^{r,\nu }} \end{aligned}$$
for \(0< t < \min \{\mu \rho ^2, T\}\), where C is a positive constant depending only on n, p and r.
Proof of Proposition 3.1
Let \(x\in \Omega\) and \(\zeta\) be a smooth function in \(C_0^{\infty }(\Omega )\) such that
$$\begin{aligned} \left\{ \begin{aligned}&0\le \zeta \le 1 \text { and } |\nabla \zeta |\le 2\rho ^{-1} \text { in } \Omega ,\\&\zeta =1 \text { on } B_\rho (x), \quad \zeta =0 \text { in } \Omega \setminus B_{2\rho }(x). \end{aligned} \right. \end{aligned}$$
For any \(\ 0<\tau <t \le T\), multiplying (1.1) by \((\text{sgn}\,u)|u|^{r-1}\zeta ^k\) and integrating it in \((0,\tau )\times \Omega\), we have that,
$$\begin{aligned} & \frac{1}{r}\int\nolimits_{{B_{{2\rho }} (x)}} {\left| {u(\tau ,y)} \right|^{r} } \zeta (y)^{k} {\text{d}}y - \frac{1}{r}\int_{{B_{{2\rho }} (x)}} {\left| {u(0,y)} \right|^{r} } \zeta (y)^{k} {\text{d}}y \\ & \quad + \int_{0}^{\tau } {\int_{\Omega } \nabla } u(s,y) \cdot \nabla (({\text{sgn}}\;u(s,y))\left| {u(s,y)} \right|^{{r - 1}} \zeta (y)^{k} ){\text{d}}y{\text{d}}s \\ & \quad = \int_{0}^{\tau } {\int_{\Omega } a } \cdot \nabla (\left| {u(s,y)} \right|^{{p - 1}} u(s,y))({\text{sgn}}\;u(s,y))\left| {u(s,y)} \right|^{{r - 1}} \zeta (y)^{k} {\text{d}}y{\text{d}}s. \\ \end{aligned}$$
(3.2)
As the relations
$$\begin{aligned} \nabla u\cdot \nabla ( (\text{sgn}\, u)|u|^{r-1}\zeta ^k) \ge&(r-1)|u|^{r-2}|\nabla u|^2 \zeta ^k- \left| k (\text{sgn}\,u)|u|^{r-1}\zeta ^{k-1}\nabla u \cdot \nabla \zeta \right| \\ \ge&\{(r-1)-k\varepsilon \}|u|^{r-2} |\nabla u|^2 \zeta ^k -\frac{k}{4\varepsilon } |u|^r\zeta ^{k-2}|\nabla \zeta |^2 \end{aligned}$$
(3.3)
and
$$\begin{aligned} \left| \nabla (|u|^{\frac{r}{2}}\zeta ^{\frac{k}{2}}) \right| ^2&\le \frac{r^2}{2}|u|^{r-2} | \nabla u|^2 \zeta ^k +\frac{k^2}{2}|u|^r\zeta ^{k-2}|\nabla \zeta |^2 \end{aligned}$$
(3.4)
are hold. Hence, by inequalities (3.3) and (3.4), we have that,
$$\begin{aligned} \nabla u\cdot \nabla&( (\text{sgn}\, u)|u|^{r-1}\zeta ^k)\\ \ge&\{(r-1)-\varepsilon k\} \left\{ \frac{2}{r^2} \left| \nabla (|u|^{\frac{r}{2}}\zeta ^{\frac{k}{2}}) \right| ^2 -\frac{k^2}{r^2}|u|^r\zeta ^{k-2}|\nabla \zeta |^2 \right\} \\&-\frac{k}{4\varepsilon } |u|^r\zeta ^{k-2}|\nabla \zeta |^2 \\ =&C_1 \left| \nabla (|u|^{\frac{r}{2}}\zeta ^{\frac{k}{2}}) \right| ^2 - C_2 |u|^r\zeta ^{k-2}|\nabla \zeta |^2. \end{aligned}$$
(3.5)
Moreover, by Young’s inequality, we have that,
$$\begin{aligned} & \int_{0}^{\tau } {\int_{{B_{{2\rho }} (x)}} {\frac{p}{{p + r - 1}}} } a \cdot \nabla \left( {\left| {u(s,y)} \right|^{{p + r - 1}} } \right)\zeta (y)^{k} {\text{d}}y{\text{d}}s \\ & = \frac{{kp}}{{p + r - 1}}\int_{0}^{\tau } {\int_{{B_{{2\rho }} (x)}} a } \cdot \left| {u(s,y)} \right|^{{p + r - 1}} \zeta (y)^{{k - 1}} \nabla \zeta (y){\text{d}}y{\text{d}}s \\ & \le C_{2} \int\limits_{0}^{\tau } {\int\limits_{{B_{{2\rho }} (x)}} {\left| {u(s,y)} \right|} ^{r} } \zeta (y)^{{k - 2}} \left| {\nabla \zeta (y)} \right|^{2} {\text{d}}y{\text{d}}s \\ & \quad + C_{3} \int_{0}^{\tau } {\int_{{B_{{2\rho }} (x)}} | } u(s,y)|^{{2p + r - 2}} \zeta (y)^{k} {\text{d}}y{\text{d}}s. \\ \end{aligned}$$
(3.6)
Hence, by inequalities (3.2), (3.5) and (3.6), we have that,
$$\begin{aligned} & \frac{1}{r}\int\nolimits_{{B_{{2\rho }} (x)}} {\left| {u(\tau ,y)} \right|} ^{r} \zeta (y)^{k} {\text{d}}y - \frac{1}{r}\int\nolimits_{{B_{{2\rho }} (x)}} {\left| {u(0,y)} \right|} ^{r} \zeta (y)^{k} {\text{d}}y \\ & \quad + C_{1} \int_{0}^{\tau } {\int_{{B_{{2\rho }} (x)}} {\left| {\nabla (\left| {u(s,y)} \right|^{{\frac{r}{2}}} \zeta (y)^{{\frac{k}{2}}} )} \right|^{2} } } {\text{d}}y{\text{d}}s \\ & \le C_{2} \int\nolimits_{0}^{\tau } {\int\nolimits_{{B_{{2\rho }} (x)}} {\left| {u(s,y)} \right|} } ^{r} \zeta (y)^{{k - 2}} \left| {\nabla \zeta (y)} \right|^{2} {\text{d}}y{\text{d}}s \\ & \quad + C_{3} \int\nolimits_{0}^{\tau } {\int\nolimits_{{B_{{2\rho }} (x)}} {\left| {u(s,y)} \right|} } ^{{2p + r - 2}} \zeta (y)^{k} {\text{d}}y{\text{d}}s. \\ \end{aligned}$$
(3.7)
We now estimate the last term of the right hand side of (3.7) using Gagliardo–Nirenberg’s inequality (Proposition 2.4). In particular, choosing \({\tilde{q}}=\frac{4}{r}(p-1)+2\) and \(\frac{{\tilde{q}}\theta }{2}=1\), and setting \(g(s,y):=|u(s,y)|\zeta (y)^{\frac{k}{2p+r-2}}\), we have using Hölder’s inequality for \(r\ge n(p-1)\) that
$$\begin{aligned} \int _0^{\tau } \int _{B_{2\rho } (x)}&|u(s,y)|^{2p+r-2}\zeta (y)^k\text{d}y\text{d}s \\ \le&C \sup _{0<s<\tau } \left( \int _{B_{2\rho } (x)}|g(s,y)|^{n(p-1)}\text{d}y \right) ^{\frac{2}{n}} \int _0^t \int _{B_{2\rho } (x)} |\nabla |g(s,y)|^{\frac{r}{2}}|^2\text{d}y\text{d}s \\ \\ \le&C \sup _{0<s<\tau } \left( \rho ^{\frac{r}{p-1}-n} \int _{B_{2\rho } (x)} |u(s,y)|^r\text{d}y \right) ^{\frac{2(p-1)}{r}}\\&\quad \times \int _0^{\tau } \left( \int _{B_{2\rho } (x)} \left| \nabla (| u(s, y)|^\frac{r}{2}) \right| ^2\text{d}y +\rho ^{-2}\int _{B_{2\rho } (x)}|u(s,y)|^r\text{d}y \right) \text{d}s. \end{aligned}$$
(3.8)
Hence, by Proposition 2.1, we conclude from (3.7) and (3.8) that
$$\begin{aligned} &\frac{1}{r}\int _{B_{2\rho }(x)} |u(\tau , y)|^r\zeta (y)^k \text{d}y - \frac{1}{r}\int _{B_{2\rho }(x)} |u(0, y)|^r\zeta (y)^k \text{d}y \\&\quad +C_1\int _0^{\tau } \int _{B_{2\rho }(x)} \left| \nabla (|u(s, y)|^{\frac{r}{2}}\zeta (y)^{\frac{k}{2}}) \right| ^2\text{d}y\text{d}s\\&\quad \le C_2\int _0^{\tau }\int _{B_{2\rho }(x)}|u(s, y)|^r \zeta (y)^{k-2} |\nabla \zeta (y)|^2 \text{d}y\text{d}s \\&\quad + C_3\sup _{0<s<\tau } \left( \rho ^{\frac{r}{p-1}-n} \int _{B_{2\rho } (x)} |u(s,y)|^r\text{d}y \right) ^{\frac{2(p-1)}{r}}\\&\quad \times \int _0^{\tau } \left( \int _{B_{2\rho } (x)} \left| \nabla (| u(s, y)|^\frac{r}{2}) \right| ^2\text{d}y+\rho ^{-2}\int _{B_{2\rho } (x)}|u(s,y)|^r\text{d}y \right) \text{d}s \\&\quad \le C_2\rho ^{-2}\int _0^{\tau }\int _{B_{\rho }(x)}|u(s,y)|^r\text{d}y\text{d}s\\&\quad + C_3\sup _{0<s<\tau } \left( \rho ^{\frac{r}{p-1}-n} \int _{B_{\rho } (x)} |u(s,y)|^r\text{d}y \right) ^{\frac{2(p-1)}{r}}\\&\quad \times \int _0^{\tau } \left( \int _{B_{\rho } (x)} \left| \nabla (| u(s, y)|^\frac{r}{2}) \right| ^2\text{d}y+\rho ^{-2}\int _{B_{\rho } (x)}|u(s,y)|^r\text{d}y \right) \text{d}s. \end{aligned}$$
(3.9)
By taking, the supremum for \(\tau \in (0,t)\) in the right hand side of (3.9) and using (3.1), we have that,
$$\begin{aligned} &\int _{B_{\rho }(x)} |u(\tau , y)|^r \text{d}y +C_1\int _0^\tau \int _{B_{\rho }(x)} \left| \nabla (|u(s, y)|^{\frac{r}{2}}) \right| ^2\text{d}y\text{d}s \\&\quad \le C_2 t\rho ^{-2} \sup _{0<s<t} \int _{B_{\rho }(x)}|u(s,y)|^r\text{d}y + \sup _{x\in \mathbb {R}^{n} } \int _{B_{\rho }(x)}|u(0, y)|^r\text{d}y \\&\quad + C_3\sup _{0<s<t} \left( \rho ^{\frac{r}{p-1}-n} \int _{B_{\rho } (x)} |u(s,y)|^r\text{d}y \right) ^{\frac{2(p-1)}{r}}\\&\quad \times \int _0^{t} \left( \int _{B_{\rho } (x)} \left| \nabla (| u(s, y)|^\frac{r}{2}) \right| ^2\text{d}y+\rho ^{-2}\int _{B_{\rho } (x)}|u(s,y)|^r\text{d}y \right) \text{d}s \\&\quad \le C_2 t\rho ^{-2} \sup _{0<s<t} \int _{B_{\rho }(x)}|u(s,y)|^r\text{d}y + \sup _{x\in \mathbb {R}^{n} } \int _{B_{\rho }(x)}|u(0, y)|^r\text{d}y \\&\quad + C_3\gamma _{1} ^{2(p-1)} \int _0^{t} \left( \int _{B_{\rho } (x)} \left| \nabla (| u(s, y)|^\frac{r}{2}) \right| ^2\text{d}y+\rho ^{-2}\int _{B_{\rho } (x)}|u(s,y)|^r\text{d}y \right) \text{d}s. \end{aligned}$$
(3.10)
for \(0<\tau <t\le T\). Taking a sufficiently small \(\gamma _1\) and \(t\rho ^{-2}\) if necessary, and by taking the
supremum for \(\tau \in (0,t)\), we deduce from (3.10) that
$$\begin{aligned} &\sup _{0<\tau<t} \int _{B_{\rho }(x)} |u(\tau ,y)|^r \text{d}y \le&C t\rho ^{-2} \sup _{0<s<t} \int _{B_{\rho }(x)}|u(s,y)|^r\text{d}y\text{d}s + \int _{B_{\rho }(x)}|u(0, y)|^r\text{d}y \end{aligned}$$
for \(0<\tau <t\le T\). This implies that,
$$\begin{aligned} \big (1-C t\rho ^{-2}\big )^{\frac{\nu }{r}}\sup _{0<s<t} \bigg (\int _{B_{\rho }(x)} |u(s,y)|^r \text{d}y\bigg )^{\frac{\nu }{r}} \le \bigg (\int _{B_{\rho }(x)}|u(0,y)|^r\text{d}y\bigg )^{\frac{\nu }{r}}. \end{aligned}$$
Taking summation on both sides on the lattice point \(x_{k}\in \mathbb {Z}^{n}\), we have that,
$$\begin{aligned} &\big (1-C t\rho ^{-2}\big )^{\frac{\nu }{r}}\sup _{0<s<t} \sum _{x_{k}\in \rho \mathbb {Z}^{n}} \bigg (\int _{B_{\rho }(x_{k})} |u(s,y)|^r \text{d}y\bigg )^{\frac{\nu }{r}}\\&\quad \le \sum _{x_{k}\in \rho \mathbb {Z}^{n}} \bigg (\int _{B_{\rho }(x_{k})}|u(0,y)|^r\text{d}y\bigg )^{\frac{\nu }{r}}. \end{aligned}$$
(3.11)
Hence, from (3.11), we have that,
$$\begin{aligned} \sup _{0<s<t} \Vert u(s)\Vert _{L_{\rho }^{r,\nu }} \le C \Vert u(0)\Vert _{ L_{\rho }^{r,\nu }} , \end{aligned}$$
for \(0< t < \min \{\mu \rho ^2, T\}\). \(\square\)
Proposition 3.2
(Difference estimate) Let r satisfy (1.3), \(r>1\) and \(T>0\). Let \(u_{0}\) and \(v_{0} \in L_{\rho }^{r,\nu }(\Omega )\) be two initial data and suppose that u and v be a corresponding \({L}^{r,\nu }(\Omega )\)- solution of (1.1) in \((0, T)\times \Omega\), respectively. There exists a positive constant \(\gamma _2\) such that, if
$$\begin{aligned} \rho ^{\frac{1}{p-1}-\frac{n}{r}} \sup _{0\le s\le T}\Vert u(s)\Vert _{L_{\rho }^{r,\nu }} \le \gamma _2,\\ \rho ^{\frac{1}{p-1}-\frac{n}{r}} \sup _{0\le s\le T}\Vert v(s)\Vert _{L_{\rho }^{r,\nu }} \le \gamma _2, \end{aligned}$$
(3.12)
for some \(\rho >0\), then there exists a constant \(\mu >0\) depending only on p, r, n and \(\gamma _2\) such that
$$\begin{aligned} \sup _{0<s<t}\Vert u(s)-v(s)\Vert _{L_{\rho }^{r,\nu }} \le C \Vert u_0-v_0\Vert _{ L_{\rho }^{r,\nu }} \end{aligned}$$
for \(0< t < \min \{\mu \rho ^2, T\}\), where C and \(\mu\) are positive constants depending only on n, p and r.
Proof of Proposition 3.2
Let \(x\in \Omega\) and \(\zeta\) be a smooth function in \(C_0^{\infty }(\Omega )\) defined in (1.1) Suppose that u and v are two strong solutions of (1.1) in \((0, T) \times \Omega\) and let \(w=u-v\). Then multiply \(|w|^{r-1}(\text{sgn}\, w) \zeta ^k\) for \(k\in \mathbb {N}\) to the difference of equation
$$\begin{aligned} \partial _t w-\Delta w = a\cdot \nabla (|u|^{p-1}u-|v|^{p-1}v) \end{aligned}$$
and integrate it over \(\Omega\) we obtain that
$$\begin{aligned} \frac{1}{r} \frac{d}{dt}&\int _{\Omega }|w(s)|^r \zeta ^k \text{d}y +\int _{\Omega } \nabla w(s)\cdot \nabla (|w(s)|^{r-1}(\text{sgn}\, w(s))\zeta ^k \text{d}y \\ =&-\int _{\Omega } (|u|^{p-1}u-|v|^{p-1}v) \big (a\cdot \nabla (\text{sgn}\, w(s)|w(s)|^{r-1}) \big ) \zeta ^k \text{d}y \\&-\int _{\Omega } (|u|^{p-1}u-|v|^{p-1}v) (\text{sgn}\,w(s))|w(s)|^{r-1}a\cdot \nabla \zeta ^k \text{d}y. \end{aligned}$$
(3.13)
Observing that
$$\begin{aligned} \nabla w\cdot \nabla&(|w|^{r-1}(\text{sgn}\, w)\zeta ^k) \ge C_1 \left| \nabla (|w|^{\frac{r}{2}}\zeta ^{\frac{k}{2}}) \right| ^2 - C_2 |w|^r\zeta ^{k-2}|\nabla \zeta |^2. \end{aligned}$$
(3.14)
By mean value’s theorem, we know that
$$\begin{aligned} \big ||u|^{p-1}u-|v|^{p-1}v\big | =&\bigg |\int _0^1 \frac{d}{d\theta } \big (|v+\theta (u-v)|^{p-1}(v+\theta (u-v))\big )d\theta \bigg | \\ \le&p|u-v|\int _0^1\bigg ( |v+\theta (u-v)|^{p-1}\bigg )d\theta \\ \le&p|w|\big (\max (|u|, |v|)\big )^{p-1}. \end{aligned}$$
(3.15)
Therefore, by (3.14) and (3.15), we obtain from (3.13) that
$$\begin{aligned} \frac{1}{r} \frac{d}{dt}&\int _{B_{2\rho }(x)}|w(s)|^r \zeta ^k \text{d}y +C\int _{B_{2\rho }(x)} \left| \nabla (|w(s)|^{\frac{r}{2}}\zeta ^{\frac{k}{2}}) \right| ^2\text{d}y\\&-C\int _{B_{2\rho }(x)}|w(s)|^r\zeta ^{k-2}|\nabla \zeta |^2 \text{d}y \\&\le C\int _{B_{2\rho }(x)} \big (\max (|u(s)|, |v(s)|)\big )^{p-1} \big | \nabla |w(s)|^{r} \big | \zeta ^k \text{d}y \\&+C \int _{B_{2\rho }(x)} \big (\max (|u(s)|, |v(s)|)\big )^{p-1} |w(s)|^{r} |\nabla \zeta ^k| \text{d}y. \end{aligned}$$
(3.16)
Now we estimate the first and last term of the right hand side of (3.16) using the Young and the Hölder inequalities. The first term of the right hand side of (3.16) follows:
Let \(U(s)=\max (|u(s)|, |v(s)|)\), then
$$\begin{aligned} &\int _{B_{2\rho }(x)} \big (\max (|u(s)|, |v(s)|)\big )^{p-1} \big | \nabla |w(s)|^{r} \big | \zeta ^k \text{d}y \\&\quad = C\int _{B_{2\rho }(x)} U(s)^{p-1}|w(s)|^{\frac{r}{2}} \big | \nabla |w(s)|^{\frac{r}{2}} \big | \zeta ^k \text{d}y \\&\quad \le C\int _{B_{2\rho }(x)} U(s)^{2p-2} |w(s)|^r \zeta ^{k}\text{d}y +C\int _{B_{2\rho }(x)} \big | \nabla |w(s)|^{\frac{r}{2}} \big |^2\zeta ^k \text{d}y. \end{aligned}$$
(3.17)
Now we estimate the first term of the right hand side of (3.17) using the Hölder and the Sobolev inequalities and obtain that
$$\begin{aligned} &\int _0^{\tau } \int _{B_{2\rho } (x)} |U(s,y)|^{2p-2}|w(s,y)|^r\zeta (y)^k\text{d}y\text{d}s \\&\quad \le C \int _0^{\tau } \left( \int _{B_{2\rho } (x)}|U(s,y)|^{n(p-1)}\text{d}y \right) ^{\frac{2}{n}} \left( \int _{B_{2\rho } (x)} \left| |w(s,y)|^{\frac{r}{2}} \right| ^\frac{2n}{n-2}\zeta (y)^\frac{kn}{n-2}\text{d}y \right) ^\frac{n-2}{n}\text{d}s \\&\quad \le C \sup _{0<s<\tau } \left( \rho ^{\frac{r}{p-1}-n} \int _{B_{2\rho } (x)}|U(s,y)|^r\text{d}y \right) ^{\frac{2(p-1)}{r}} \\&\quad \times \int _0^{\tau } \left( \int _{B_{2\rho } (x)} \left| \nabla (| w(s, y)|^\frac{r}{2}) \right| ^2\text{d}y+\rho ^{-2}\int _{B_{2\rho } (x)}|w(s,y)|^r\text{d}y \right) \text{d}s. \end{aligned}$$
(3.18)
Therefore, by (3.17), (3.18), we obtain from (3.16) that
$$\begin{aligned} \frac{1}{r}\int _{B_{2\rho }(x)}&|w(\tau , y)|^r\zeta (y)^k \text{d}y - \frac{1}{r}\int _{B_{2\rho }(x)} |w(0, y)|^r\zeta (y)^k \text{d}y \\&+C_{1}\int _0^{\tau } \int _{B_{2\rho }(x)} \left| \nabla (|w(s, y)|^{\frac{r}{2}}\zeta (y)^{\frac{k}{2}}) \right| ^2\text{d}y\text{d}s\\ \le&C_{2}\int _0^{\tau }\int _{B_{2\rho }(x)}|w(s, y)|^r \zeta (y)^{k-2} |\nabla \zeta (y)|^2 \text{d}y\text{d}s \\&+ C_{3} \sup _{0<s<\tau } \left( \rho ^{\frac{r}{p-1}-n} \int _{B_{2\rho } (x)}\big | \max (|u|, |v|)\big |^r\text{d}y \right) ^{\frac{2(p-1)}{r}} \\&\quad \times \int _0^{\tau } \left( \int _{B_{2\rho } (x)} \left| \nabla (| w(s, y)|^\frac{r}{2}) \right| ^2\text{d}y+\rho ^{-2}\int _{B_{2\rho } (x)}|w(s,y)|^r\text{d}y \right) \text{d}s. \end{aligned}$$
(3.19)
By the Gagliardo–Nirenberg inequality, we obtain from (3.19) that
$$\begin{aligned} &\frac{1}{r}\int _{B_{2\rho }(x)} |w(\tau , y)|^r\zeta (y)^k \text{d}y - \frac{1}{r}\int _{B_{2\rho }(x)} |w(0, y)|^r\zeta (y)^k \text{d}y \\&\quad +C_{1}\int _0^\tau \int _{B_{2\rho }(x)} \left| \nabla (|w(s,y)|^{\frac{r}{2}} \zeta (y)^{\frac{k}{2}}) \right| ^2\text{d}y\text{d}s\\&\quad \le C_{2}\int _0^\tau \int _{B_{\rho }(x)}|w(s,y)|^r\text{d}y\text{d}s\\&\quad +C_{3} \left( \rho ^{\frac{r}{p-1}-n} \sup _{0<s<\tau } \int _{B_{\rho } (x)}\big ||u(s,y)|+|v(s,y)|\big |^r\text{d}y \right) ^{\frac{2(p-1)}{r}} \\&\quad \times \int _0^{\tau } \left( \int _{B_{\rho } (x)} \left| \nabla (| w(s,y)|^\frac{r}{2}) \right| ^2\text{d}y+\rho ^{-2}\int _{B_{\rho } (x)}|w(s,y)|^r\text{d}y \right) \text{d}s \\&\quad \le C_{2}\rho ^{-2} \int _0^\tau \int _{B_{\rho }(x)}|w(s,y)|^r\text{d}y\text{d}s\\&\quad +C_{3} \rho ^{\frac{r}{p-1}-n} \sup _{0<s<\tau } \left( \int _{B_{\rho } (x)}|u(s,y)|^{r}\text{d}y+ \int _{B_{\rho } (x)}|v(s,y)|^r\text{d}y\right) ^{\frac{2(p-1)}{r}} \\&\quad \times \int _0^{\tau } \left( \int _{B_{\rho } (x)} \left| \nabla (| w(s, y)|^\frac{r}{2}) \right| ^2\text{d}y+\rho ^{-2}\int _{B_{\rho } (x)}|w(s,y)|^r\text{d}y \right) \text{d}s. \end{aligned}$$
(3.20)
for all \(0<\tau <t\le T\).
By taking the supremum for \(\tau \in (0,t)\) in the right hand side of (3.20) and using (3.12), we obtain that
$$\begin{aligned} &\int _{B_{\rho }(x)} |w(\tau , y)|^r \text{d}y +C_{1}\int _0^\tau \int _{B_{\rho }(x)} \left| \nabla (|w(s, y)|^{\frac{r}{2}}) \right| ^2\text{d}y\text{d}s \\&\quad \le C_{2}t\rho ^{-2} \sup _{0<s<t} \int _{B_{\rho }(x)}|w(s,y)|^r\text{d}y +\int _{B_{\rho }(x)}|w(0, y)|^r\text{d}y \\&\quad +C_{3}\gamma _2 ^{2(p-1)} \left( \int _0^{t} \int _{B_{\rho } (x)} \left| \nabla (| w(s, y)|^\frac{r}{2}) \right| ^2\text{d}y\text{d}s+t\rho ^{-2}\sup _{0<s<t}\int _{B_{\rho } (x)}|w(s,y)|^r\text{d}y \right) . \end{aligned}$$
(3.21)
for \(0<\tau <t\le T\). Taking a sufficiently small \(\gamma _2\) and \(t\rho ^{-2}\) if necessary, and by taking the supremum for \(\tau \in (0,t)\), we deduce from (3.21) that
$$\begin{aligned} &\sup _{0<\tau<t} \int _{B_{\rho }(x)} |w(\tau ,y)|^r \text{d}y \le&C t\rho ^{-2} \sup _{0<s<t} \int _{B_{\rho }(x)}|w(s,y)|^r\text{d}y\text{d}s + \int _{B_{\rho }(x)}|w(0, y)|^r\text{d}y \end{aligned}$$
for \(0<\tau <t\le T\). This implies that
$$\begin{aligned} \big (1-C t\rho ^{-2}\big )^{\frac{\nu }{r}}\sup _{0<s<t} \bigg (\int _{B_{\rho }(x)} |w(s,y)|^r \text{d}y\bigg )^{\frac{\nu }{r}} \le \bigg (\int _{B_{\rho }(x)}|w(0,y)|^r\text{d}y\bigg )^{\frac{\nu }{r}}. \end{aligned}$$
Taking summation on both sides on the lattice point \(x_{k}\in \mathbb {Z}^{n},\) we have
$$\begin{aligned} \big (1-C t\rho ^{-2}\big )^{\frac{\nu }{r}}\sup _{0<s<t} \sum _{x_{k}\in \rho \mathbb {Z}^{n}} \bigg (\int _{B_{\rho }(x_{k})} |w(s,y)|^r \text{d}y\bigg )^{\frac{\nu }{r}} \le \sum _{x_{k}\in \rho \mathbb {Z}^{n}} \bigg (\int _{B_{\rho }(x_{k})}|w(0,y)|^r\text{d}y\bigg )^{\frac{\nu }{r}}. \end{aligned}$$
(3.22)
Hence, from (3.22), we obtain that
$$\begin{aligned} \sup _{0<s<t} \Vert w(s)\Vert _{L_{\rho }^{r,\nu }} \le C \Vert w(0)\Vert _{ L_{\rho }^{r,\nu }} ,\end{aligned}$$
for \(0< t < \min \{\mu \rho ^2, T\}\). \(\square\)
To obtain the critical existence of the weak solutions, the \(L^{\infty }\) a priori estimate for the weak solutions is essential. For related results, see ([1, 24]).
Proposition 3.3
(\(L^{\infty }\)-a priori estimate) Let u be a \(L^{r,\nu }(\Omega )\)-solution of (1.1) in \((0, T)\times \Omega\), where \(0<T<\infty\) and \(r>1\). For some positive constant \(\gamma _3\), if
$$\begin{aligned} \rho ^{\frac{1}{p-1}-\frac{n}{r}} \sup _{0\le s\le T}\Vert u(s)\Vert _{L_{\rho }^{r,\nu }} \le \gamma _3 \end{aligned}$$
(3.23)
for some \(\rho >0\), then there exists a constant \(C>0\) such that
$$\begin{aligned} \Vert u\Vert _{L^{\infty }{\big ((t_1, t)\times B_{R_1}(x)}\big )} \le CD^{\frac{n+2}{2r}}\bigg (\int _{t_2}^t\int _{B_{R_2}(x)}|u|^r\text{d}y\text{d}s\bigg )^{\frac{1}{r}}, \end{aligned}$$
(3.24)
$$\begin{aligned} \int _{t_1}^t\int _{B_{R_1}(x)}|\nabla u|^2\text{d}y\text{d}s \le CD\int _{t_2}^t\int _{B_{R_2}(x)}|u|^2\text{d}y\text{d}s, \end{aligned}$$
(3.25)
for all \(x\in \Omega\), \(0<R_1<R_2\) and \(0<t_2<t_1\le T\), where
$$\begin{aligned} D=C_1(R_2-R_1)^{-2}+(t_1-t_2)^{-1}. \end{aligned}$$
Proof of Proposition 3.3
Let \(x \in \Omega\), \(0< R_1<R_2\), \(0<t_2<t_1<t \le T\). For \(j=0,1,2,...\), set
$$\begin{aligned} r_j:=R_1+(R_2-R_1)2^{-j}, \quad \tau _j:=t_1-(t_1-t_2)2^{ -2j}, \quad Q_j=( \tau _j, t) \times B_{r_j} (x). \end{aligned}$$
Let \(\zeta _j\) be a piecewise smooth function in \(Q_j\) satisfying
$$\begin{aligned} \left\{ \begin{aligned}&0 \le \zeta _j(t,x)\le 1 \quad \text { in }\Omega ,\\&\quad \zeta _j(t,x)\equiv 1 \quad \text { on } Q_{j+1},\\&\zeta _j=0 \quad \text { near }\quad [\tau _j, t]\times \partial B_{r_j}(x) \cup \{\tau _j\}\times B_{r_j}(x),\\&|\nabla \zeta _j|\le \frac{2^{j+1}}{R_2-R_1}, \quad \text { in } \ Q_j\\&0\le \partial _t \zeta _j\le \frac{2^{2(j+1)}}{t_2-t_1} \quad \text { in } \ Q_j. \end{aligned} \right. \end{aligned}$$
(3.26)
Multiplying (1.1) by \(|u(t,y)|^{\beta -2}u(t,y)\zeta ^k(t,y)\) and integrating it in \(\Omega\), we obtain that
$$\begin{aligned} &\frac{1}{\beta }\frac{d}{dt} \bigg ( \int _{B_{r_j}(x)} |u(t,y)|^{\beta }\zeta _j(t,y)^k \text{d}y \bigg ) +\frac{2(2\beta -k-2)}{\beta ^2}\int _{B_{r_j}(x)} \left| \nabla |u(t,y)|^{\frac{\beta }{2}} \right| ^2 \zeta _j(t,y)^k\text{d}y\\&\quad \le \frac{pk|a|^2}{2(p+\beta -1)} \int _{B_{r_j}(x)}|u(t,y)|^{2p+\beta -2} \zeta _j(t,y)^k\text{d}y \\&\quad +\frac{k}{2}\bigg (\frac{p}{p+\beta -1}+1\bigg ) \int _{B_{r_j}(x)}|u(t,y)|^\beta \zeta _j(t,y)^{k-2} |\nabla \zeta _j(t,y)|^2\text{d}y\\&\quad + \frac{k}{\beta }\int _{B_{r_j}(x)} \zeta _j(t,y)^{k-1} |u(t,y)|^\beta \partial _t\zeta _j(t,y) \text{d}y. \end{aligned}$$
(3.27)
For the highest-order term, using the Hölder and the Sobolev inequalities, we obtain that
$$\begin{aligned} \int _{B_{r_j}(x)}&|u(t,y)|^{2(p-1)}|u(t,y)|^{\beta }{\zeta _j}^k\text{d}y \\ \le&\bigg (\int _{B_{r_j}(x)}|u(t,y)|^{n(p-1)}\text{d}y\bigg )^{\frac{2}{n}} \bigg ( \int _{B_{r_j}(x)} \big ( |u(t,y)|^{\beta }{\zeta _j}^k \big )^\frac{n}{n-2}\text{d}y \bigg )^{\frac{n-2}{n}} \\ \le&C_s^2 \bigg (\int _{B_{r_j}(x)}|u(t,y)|^{n(p-1)}\text{d}y\bigg )^{\frac{2}{n}} \bigg ( \int _{B_{r_j}(x)} \big |\nabla \big ( |u(t,y)|\zeta _j(t,y)^{\frac{k}{\beta }} \big )^{\frac{\beta }{2}} \big |^2 \text{d}y \bigg ). \end{aligned}$$
(3.28)
Since
$$\begin{aligned} \frac{1}{2} \left| \nabla ( u{\zeta _j}^\frac{k}{\beta })^{\frac{\beta }{2}} \right| ^2-\frac{k^2}{4}u^\beta {\zeta _j}^{k-2}|\nabla {\zeta _j} |^2 \le \left| \nabla u^{\frac{\beta }{2}} \right| ^2{\zeta _j}^k \end{aligned}$$
and using (3.28), integrating (3.27) over \(t\in I_j\), we obtain that
$$\begin{aligned} \sup _{t\in I_j}&\int _{B_{r_j}(x)} |u(s,y)|^{\beta } \zeta _j(s,y)^k \text{d}y +\frac{2\beta -k-2}{\beta }\int _{I_j}\int _{B_{r_j}(x)} \left| \nabla ( |u(s,y)| \zeta _j(s,y)^\frac{k}{\beta } )^{\frac{\beta }{2}} \right| ^2\text{d}y\text{d}s \\ \le&\frac{pk|a|^2\beta }{2(p+\beta -1)} C_s^2\int _{I_j}\bigg \{ \bigg ( \int _{B_{r_j}(x)}|u(s,y)|^{n(p-1)}\text{d}y \bigg )^{\frac{2}{n}} \int _{B_{r_j}(x)} \big |\nabla \big (|u(s,y)|\zeta _j(s,y)^{\frac{k}{\beta }} \big )^{\frac{\beta }{2}} \big |^2 \text{d}y\bigg \}\text{d}s \\&+\frac{k}{2}\bigg (\frac{p\beta }{p+\beta -1} +\beta +\frac{k(2\beta -k-2)}{\beta } \bigg ) \int _{I_j}\int _{B_{r_j}(x)} |u(s,y)|^{\beta } \zeta _j(s,y)^{k-2} |\nabla \zeta _j(s,y) |^2\text{d}y\\&\quad + k\int _{I_j}\int _{B_{r_j}(x)} |u(s,y)|^{\beta } \zeta _j^{k-1}(t,y) \partial _t\zeta _j(s,y) \text{d}y. \end{aligned}$$
(3.29)
Let \(\gamma _3>0\) be taken as
$$\begin{aligned} \frac{pk|a|^2\beta }{2(p+\beta -1)} C_s^2\gamma _3^{\frac{p-1}{2}} \le 1-\frac{k+2}{n(p-1)}. \end{aligned}$$
Then, under the assumption (3.23), we estimate the first term of the right hand side of (3.29) and it cancels by the second term of the right hand side. Thus from (3.29) and using the estimate for the derivatives \(\zeta _j\) in (3.26), that
$$\begin{aligned} \sup _{t\in I_j}&\int _{B_{r_j}(x)} u(s)^\beta {\zeta _j(s)}^k \text{d}y +\int _{I_j}\int _{B_{r_j}(x)} \left| \nabla ( u{\zeta _j}^\frac{k}{\beta })^{\frac{\beta }{2}} \right| ^2\text{d}y\text{d}s\\ \le&2k\bigg [ \bigg ( \frac{p\beta }{p+\beta -1} +\beta +\frac{k(2\beta -k-2)}{\beta } \bigg )\frac{2^{2j}}{(R_2-R_1)^2} +\frac{2^{2j}}{t_1-t_2} \bigg ] \\&\int _{I_j} \int _{B_{r_j}(x)}|u(s,y)|^{\beta } \text{d}y\text{d}s\\ =&C{2^{2j}} \bigg \{\frac{\beta }{(R_2-R_1)^2} +\frac{1}{t_1-t_2} \bigg \} \int _{I_j} \int _{B_{r_j}(x)}|u(s,y)|^{\beta } \text{d}y\text{d}s, \end{aligned}$$
(3.30)
for any \(j=0,1,2,...\) and \(\beta >r\). Now applying the Gagliardo–Nirenberg inequality, Proposition 2.4, for any function \(f\in C_0^1({B_{r_j}(x)})\) and \(\theta \in (0, 1)\) with choosing \(r=2+\frac{4}{n}=2(1+\frac{2}{n})\), \(p=2\), \(q=2\). we obtain for letting \(\gamma =1+\frac{2}{n}\)
$$\begin{aligned} \int _{B_{r_j}(x)}|f|^{2\gamma }\text{d}y \le C^{2\gamma }\bigg (\int _{B_{r_j}(x)}|f|^2\text{d}y\bigg )^\frac{2}{n}\int _{B_{r_j}(x)}|\nabla f|^2\text{d}y. \end{aligned}$$
(3.31)
Integrating (3.31) with respect to time \(t\in I_j\), we have
$$\begin{aligned} \int _{I_j}\int _{B_{r_j}(x)}|u|^{\beta \gamma }\zeta _j^{ k\gamma }\text{d}y\text{d}s \le C^{2\gamma }\bigg (\sup _{t\in I_j}\int _{B_{r_j}(x)}|u^\beta \zeta _j^{ k}|\text{d}y\bigg )^\frac{2}{n} \int _{I_j}\int _{B_{r_j}(x)}|\nabla (u\zeta _j^{\frac{k}{\beta }})^\frac{\beta }{2}|^2\text{d}y\text{d}s. \end{aligned}$$
Hence, we obtain the reversed Hölder estimate:
$$\begin{aligned} \bigg ( \int \int _{Q_{j+1}}&|u(t,y)|^{\beta \gamma }\text{d}y\text{d}s \bigg )^{\frac{1}{\gamma }}\\ \le&\left( \int _{I_j}\int _{B_{r_j}(x)} |u(t,y)|^{\beta \gamma } \zeta _j(t,y)^{ k\gamma }\text{d}y\text{d}s \right) ^{\frac{1}{\gamma }}\\ \le&C{ 2^{2j}} \bigg \{\frac{\beta }{(R_2-R_1)^2} +\frac{1}{t_1-t_2} \bigg \} \int _{I_j} \int _{B_{r_j}(x)}|u(t,y)|^{\beta } \text{d}y\text{d}s, \end{aligned}$$
(3.32)
where \(Q_j=I_j\times B_{r_j}(x)=(\tau _j, t)\times B_{r_j}(x)\) and \(\zeta _j=1\) on \(Q_{j+1}\). Furthermore, by (3.30) with \(\beta =2\) and \(k=2\) we have (3.25). We use the estimate (3.32) iteratively with choosing
\(\beta =\beta _j=r\gamma ^j\), where \(\gamma =1+\frac{2}{n}\) and \(j=1,2,\cdots\). Since it holds
$$\begin{aligned} \bigg ( \int \int _{Q_{j+1}}&|u(t,y)|^{\beta _j\gamma }\text{d}y\text{d}s \bigg )^\frac{1}{\gamma \beta _j} \\&\le \big (C2^{2j}\big )^{\frac{1}{\beta _j}} \bigg [\frac{\beta _j}{(R_2-R_1)^2} +\frac{1}{t_1-t_2} \bigg ]^{\frac{1}{\beta _j}} \bigg (\int \int _{Q_j}|u(t,y)|^{\beta _j}\text{d}y\text{d}s \bigg )^{\frac{1}{\beta _j}}, \end{aligned}$$
we see that
$$\begin{aligned} M_{j+1} \le&\big (C2^{2j}\big )^{\frac{1}{\beta _j}} \bigg [\frac{r\gamma ^j}{(R_2-R_1)^2} +\frac{1}{t_1-t_2} \bigg ]^{\frac{1}{\beta _j}}M_j\\ =&C^{\frac{j}{\beta _j}}(CD)^{\frac{1}{\beta _j}}M_j, \end{aligned}$$
(3.33)
where
$$\begin{aligned} D=C_1(R_2-R_1)^{-2}+(t_1-t_2)^{-1}. \end{aligned}$$
The inequality (3.33) implies that
$$\begin{aligned} M_{j+1}\le M_0\prod _{k=0}^jC^{\frac{k}{\beta _k}}(CD)^{\frac{1}{\beta _k}}. \end{aligned}$$
This follows that
$$\begin{aligned} \lim _{j\rightarrow \infty }M_j \le C^{\sum _{j=0}^\infty \frac{j}{\beta _j}}(CD)^{\sum _{j=0}^\infty \frac{1}{\beta _j}}M_0. \end{aligned}$$
(3.34)
Since \(\gamma =1+\frac{2}{n}\),
$$\begin{aligned} \sum _{j=0}^\infty \frac{1}{\beta _j} \equiv \sum _{j=0}^\infty \frac{1}{r\gamma ^j} =\frac{\gamma }{r(\gamma -1)} =\frac{n+2}{2r} \end{aligned}$$
and \(\displaystyle \sum _{j=0}^\infty \frac{j}{\beta _j} < \infty .\) We obtain from (3.34)
$$\begin{aligned} \Vert u\Vert _{L^\infty (Q_\infty )} \le CD^{\frac{n+2}{2r}}\Vert u\Vert _{L^r(Q_0)}. \end{aligned}$$
Hence, we have that
$$\begin{aligned} \Vert u\Vert _{L^{\infty } {\big ((t_1, t)\times B_{R_1}(x)}\big )} \le CD^{\frac{n+2}{2r}} \bigg ( \int _{t_2}^t\int _{B_{R_2}(x)}|u|^r\text{d}y\text{d}s \bigg )^{\frac{1}{r}}. \end{aligned}$$
\(\square\)